Solve Using the Quadratic Formula -x^2+6x-5=0

$- x^{2} + 6 x - 5 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = - 1$ , $b = 6$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (- 1 \cdot - 5)}}{2 \cdot - 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot - 1 \cdot - 5}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot - 5$ .

Multiply $- 4$ by $- 1$ .

$x = \frac{- 6 \pm \sqrt{36 + 4 \cdot - 5}}{2 \cdot - 1}$

Multiply $4$ by $- 5$ .

$x = \frac{- 6 \pm \sqrt{36 - 20}}{2 \cdot - 1}$

Subtract $20$ from $36$ .

$x = \frac{- 6 \pm \sqrt{16}}{2 \cdot - 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 6 \pm \sqrt{4^{2}}}{2 \cdot - 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 6 \pm 4}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 6 \pm 4}{- 2}$

Simplify $\frac{- 6 \pm 4}{- 2}$ .

$x = 3 \pm 2$

Step 4

The final answer is the combination of both solutions.

$x = 5, 1$